How do you solve #sin(2x) - cos(x) - 2sin(x) + 1 = 0#?

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Answer 1 · 2018-04-18T00:25:15

#sin(2x) - cos(x) - 2sin(x) + 1 = 0#

#=>2sin(x)cos(x) - cos(x) - 2sin(x) + 1 = 0#

#=>cos(x)(2sin(x) - 1) -( 2sin(x) - 1) = 0#

#=>(2sin(x) - 1)( cos(x) - 1) = 0#

When #2sin(x)-1=0#

#=>sin(x)=1/2=sin(pi/6)#

#=>x=npi+(-)^npi/6" where "n inZZ#

Again when

#cos(x)=1#

#=>x=2kpi" where " k inZZ#